Damped Motion on an Inclined Surface

Derive and solve the equation of motion of the block sliding on an inclined, lubricated surface (Figure 4.4.8a). Assume that the damping force is linear. For this application the damping coefficient c depends on the contact area of the block, the viscosity of the lubricating fluid, and the thickness of the fluid layer.

Question

Damped Motion on an Inclined Surface

Derive and solve the equation of motion of the block sliding on an inclined, lubricated surface (Figure 4.4.8a). Assume that the damping force is linear. For this application the damping coefficient c depends on the contact area of the block, the viscosity of the lubricating fluid, and the thickness of the fluid layer.

Accepted Answer

We define v to be the velocity of the block parallel to the surface. The free body diagram, displaying only the forces parallel to the surface, is shown in Figure 4.4.8b. Note that the direction of the damping force must be opposite that of the velocity. The equation of motion is
[latex]m\dot{v} = mg \sin θ − cv [/latex]
Because the gravity force mg sin θ is constant, the solution is
[latex]v(t) = \left[ v(0) − \frac{mg \sin θ}{c} \right] e^{−ct/m} + \frac{mg \sin θ}{c} [/latex]
Eventually the block will reach a constant velocity of mg sin θ/c regardless of its initial velocity.
At this velocity the damping force is great enough to equal the gravity force component. Because [latex]e^{−4} = 0.02 [/latex], when t = 4m/c the velocity will be approximately 98% of its final velocity. Thus a system with a larger mass or less resistance (a smaller damping constant c) will require more time to reach its constant velocity. This apparently unrealistic result is explained by noting that the constant velocity attained is [latex]mg \sin θ/c[/latex]. Thus a system with a larger mass or a smaller c value will attain a higher velocity, and thus should take longer to reach it.

Question 4.4.1: Damped Motion on an Inclined Surface Derive and solve the eq...

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